2

ID、価格、SUM(価格) を取得したい場合は、次のようになります。

ID    price    sum
--------------------
 1     10      
 1     10      20
 2     20      
 2     20      40
 3     30
 3     30      60

これを達成する方法は何ですか?私の本当の質問:

SELECT users.login, projects.name, time_entries.issue_id, time_entries.hours
FROM users, time_entries
INNER JOIN projects ON time_entries.project_id = projects.id
WHERE time_entries.spent_on = CURDATE() - 1
AND time_entries.user_id = users.id
ORDER BY users.login;

ショー

+-------+----------------------------------------------------+----------+-------+
| login | name                                               | issue_id | hours |
+-------+----------------------------------------------------+----------+-------+
| ach   | name1                                              |    12624 |     8 |
| aco   | name2                                              |    11550 |     3 |
| aco   | name2                                              |    11585 |     3 |
| alt   | name3                                              |    12644 |   7.5 |
| ata   | name4                                              |    12761 |     1 |
| ata   | name5                                              |     NULL |     1 |
| ata   | name6                                              |    12790 |   0.5 |
| ata   | name7                                              |    12677 |   5.5 |
| ato   | name8                                              |    12530 |     8 |
| elb   | name8                                              |    12697 |     1 |
| elb   | Software management                                |    12678 |     7 |

一意のログインの 5 番目の列で時間の合計を取得したいと考えています。

4

3 に答える 3

0
SELECT users.login, projects.name, time_entries.issue_id, SUM(time_entries.hours) as hours
FROM users, time_entries
INNER JOIN projects ON time_entries.project_id = projects.id
WHERE time_entries.spent_on = CURDATE() - 1
AND time_entries.user_id = users.id GROUP BY users.login
ORDER BY users.login;
于 2013-09-26T14:28:15.910 に答える